Marie distributes toys for toddlers. She makes visits to households and gives away one toy only on visits for which the door is answered and a toddler is in residence. On any visit, the probability of the door being answered is 3/4, and the probability that there is a toddler in residence is 1/3. Assume that the events “Door answered" and “Toddler in residence" are independent and also that events related to different households are independent.
1. What is the probability that she has not distributed any toys by the end of her second visit?
2. What is the probability that she gives away the first toy on her fourth visit?
3. Given that she has given away her second toy on her fifth visit, what is the conditional probability that she will give away her third toy on her eighth visit?
Ans = 0.140
4. What is the probability that she will give away the second toy on her fourth visit?
5. Given that she has not given away her second toy by her third visit, what is the conditional probability that she will give away her second toy on her fifth visit?
6. We will say that Marie “needs a new supply"" immediately after the visit on which she gives away her last toy. If she starts out with three toys, what is the probability that she completes at least five visits before she needs a new supply?
7. If she starts out with exactly six toys, what is the expected value of the number of houses with toddlers that Marie visits without leaving any toys (because the door was not answered) before she needs a new supply?
1. 9/16
2. 27/256
3. 9/64
4. 27/256
5. 1/8
6. 243/256
7. 2
To solve these probability problems, we will need to use the principles of probability and apply them to the given information. We will also use some basic probability formulas. Let's go through each question and explain how to answer them:
1. To find the probability that Marie has not distributed any toys by the end of her second visit, we need to calculate the probability that the door is not answered and there is no toddler in residence on both visits. Since the events are independent, we can multiply the probabilities together. The probability that the door is not answered is 1 - (3/4) = 1/4, and the probability that there is no toddler in residence is 1 - (1/3) = 2/3. So, the probability that she has not distributed any toys by the end of her second visit is (1/4) * (2/3) = 1/6.
2. To find the probability that she gives away the first toy on her fourth visit, we need to calculate the probability that the door is answered on the first three visits and there is no toddler in residence on the first three visits, and then the door is answered and there is a toddler in residence on the fourth visit. Since the events are independent, we can multiply the probabilities together. The probability that the door is answered is 3/4, and the probability that there is a toddler in residence is 1/3. So, the probability that she gives away the first toy on her fourth visit is (3/4)^3 * (1/3) = 9/64.
3. To find the conditional probability that she will give away her third toy on her eighth visit, given that she has given away her second toy on her fifth visit, we need to calculate the probability that the door is answered on the next three visits and there is a toddler in residence on the next three visits, given that the door was answered and there was a toddler in residence on the first five visits. Since the events are independent, we can multiply the probabilities together. The probability that the door is answered is 3/4, and the probability that there is a toddler in residence is 1/3. So, the conditional probability is (3/4)^3 * (1/3) = 27/256.
4. To find the probability that she will give away the second toy on her fourth visit, we need to calculate the probability that the door is answered on the first three visits and there is no toddler in residence on the first three visits, and then the door is answered and there is a toddler in residence on the fourth visit, given that she has not given away the first toy on her first three visits. Since the events are independent, we can multiply the probabilities together. The probability that the door is answered is 3/4, and the probability that there is a toddler in residence is 1/3. So, the probability that she gives away the second toy on her fourth visit is (3/4)^3 * (1/3) = 9/64.
5. To find the conditional probability that she will give away her second toy on her fifth visit, given that she has not given away her second toy by her third visit, we need to calculate the probability that the door is answered on the next two visits and there is a toddler in residence on the next two visits, given that the door was answered and there was a toddler in residence on the first three visits. Since the events are independent, we can multiply the probabilities together. The probability that the door is answered is 3/4, and the probability that there is a toddler in residence is 1/3. So, the conditional probability is (3/4)^2 * (1/3) = 9/64.
6. To find the probability that Marie completes at least five visits before she needs a new supply, we need to calculate the probability that she distributes one toy on each of the first four visits, and then she either distributes one toy on the fifth visit or doesn't distribute any toy on the fifth visit. We can calculate the probability for both cases separately and then add them together. The probability that she distributes one toy on each of the first four visits is (3/4)^4 * (1/3)^4. The probability that she doesn't distribute any toy on the fifth visit is (1 - (3/4) * (1/3))^2. So, the probability that she completes at least five visits before she needs a new supply is [(3/4)^4 * (1/3)^4] + [(1 - (3/4) * (1/3))^2] = 382/729.
7. To find the expected value of the number of houses with toddlers that Marie visits without leaving any toys before she needs a new supply, we need to calculate the probability of visiting each number of houses from 0 to infinity without leaving any toys, given that she starts with exactly six toys. We can calculate the probability for each case separately and multiply it by the number of houses visited for that case, and then sum them all together. The probability of visiting 0 houses without leaving any toys is (1 - (3/4) * (1/3))^6. The probability of visiting 1 house without leaving any toys is (3/4) * (1/3) * (1 - (3/4) * (1/3))^5. Similarly, we can calculate the probabilities for visiting 2, 3, and so on houses without leaving any toys. We can stop calculating after a certain number of houses where the probability becomes negligible. Then, we can multiply each probability by the number of houses visited and sum them all together. This will give us the expected value.
These are the explanations for how to get the answers to the given probability questions. Let me know if you need any further assistance!